Question

An urn contains 6 white and 9 black balls. Two successive draws of 4 balls are made without replacement. The probability that the first draw gives all white balls, and the second draw gives all black balls, is:

• A. \( \frac{5}{256} \)
• B. \( \frac{2}{715} \)
• C. \( \frac{3}{715} \)
• D. \( \frac{3}{256} \)

Answer 1

The Correct Option is C

The objective is to determine the probability of drawing four white balls in the first draw and four black balls in the second draw, without replacement. This requires applying combinatorial probability principles.

Urn composition:

• White balls: 6
• Black balls: 9
• Total balls: 15

The process involves two consecutive draws of 4 balls each. The calculation proceeds as follows:

1. First Draw (all white balls):
   • The number of combinations for selecting 4 white balls from 6 is calculated as: \(\binom{6}{4}\).
   • \(\binom{6}{4} = 15\).
2. Total possible combinations for drawing any 4 balls from 15:
   • \(\binom{15}{4} = 1365\).
3. Probability of the first draw yielding all white balls:
   • \(\frac{\binom{6}{4}}{\binom{15}{4}} = \frac{15}{1365}\).
4. Second Draw (all black balls):
   • Following the first draw, 4 white balls are removed. The remaining balls are 2 white and 9 black.
   • The number of combinations for selecting 4 black balls from the remaining 9 is: \(\binom{9}{4}\).
   • \(\binom{9}{4} = 126\).
5. Total possible combinations for drawing any 4 balls from the remaining 11:
   • \(\binom{11}{4} = 330\).
6. Probability of the second draw yielding all black balls:
   • \(\frac{\binom{9}{4}}{\binom{11}{4}} = \frac{126}{330}\).
7. Combined Probability:
   • The probabilities of these independent events are multiplied: \(\frac{15}{1365} \times \frac{126}{330}\).
   • This simplifies to: \(\frac{15 \times 126}{1365 \times 330}\).
   • The final simplified probability is: \(\frac{3}{715}\).

The resultant probability is \(\frac{3}{715}\).